Question

Calculate the value of \[sinh[\dfrac{\pi}{6}]i\].
A. \[ - \dfrac{i}{2}\]
B. \[\dfrac{i}{2}\]
C. \[\dfrac{{i\sqrt 3 }}{2}\]
D. \[ - \dfrac{{i\sqrt 3 }}{2}\]

Answer 1

Hint: The given problem is related to hyperbolic function. This is correlated with complex numbers and trigonometric terms. First, we use the formula \[\sinh (iz) = i\sin z\] where z is a complex number.

Formula used: \[\sinh (iz) = i\sin z\] where z is a complex number. Particularly, if z is a real number then the formula is also valid.

Complete step-by-step solution:
The problem is related to hyperbolic function. This is correlated with complex numbers and trigonometric terms. First, we use the formula \[\sinh (iz) = i\sin z\] where z is a complex number.
 \[\begin{array}{l}\sinh [\dfrac{\pi }{6}i]\\ = i\sin \dfrac{\pi }{6}\\ = i.\dfrac{1}{2}\\ = \dfrac{i}{2}\end{array}\] [\[\sinh (iz) = i\sin z\]]
Hence, the option B. is correct.

Note: The course: hyperbolic function is taken from the syllabus of B.Sc. Mathematics Hons. Syllabus. The hyperbolic function is elaborately described there. Students can follow the book “Higher Algebra (Classical)” by S. K. Mapa . Students often fail to use the definition of the hyperbolic function. They directly follow the trigonometric formula. So, they should be very careful about that. Alternatively, we can use a different formula \[\sinh z = \dfrac{{\exp (z) - \exp ( - z)}}{2}\] where z is a complex number. This process takes more time. This is actually a direct formula for a hyperbolic function. In both cases, you will reach your desired solution. So, students! Don’t be puzzled. Please follow the explanation of the hyperbolic function that I have said. You should remember that \[\cosh z\] and \[\sinh z\] are periodic functions of period\[2\pi i\], where z is a complex number. In trigonometry, \[\sin x\] and \[\cos x\] are periodic functions of \[2\pi \]. So, trigonometric functions and hyperbolic functions are different. The hyperbolic function is the combination of a trigonometric function and a complex function.